Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to understand why, if $S$ is a Seifert fibered space, then $\pi_1(S)$ is residually finite. From theorems 12.2 and 11.10 in Hempel's "3-manifolds", we can work with a finite-sheeted covering $M$ such that $M$ is an $S^1$-bundle over a closed surface $T$. According to a 1987 paper by Hempel, there exists a finite-sheeted covering $\widehat{M}$ of $M$, such that $\widehat{M}$ is of the type $S^1\times F$, $F$ a closed surface. Of course, the result follows from this, but my question is:

Why does $\widehat{M}$, with such a homeomorphism type, exist?

I'm not actually sure if the existence of $\widehat{M}$ is necessary to deduce the r.f. of $\pi_1(M)$, but I don't think knowing only that it fits into the sequence $1\rightarrow\mathbb{Z}\rightarrow\pi_1(M)\rightarrow\pi_1(T)\rightarrow 1$ is enough.

share|cite|improve this question
Should note I am ignoring the special cases when $\pi_1(S)$ is virtually cyclic, since those are easy. – user641 Apr 11 '11 at 2:34
up vote 3 down vote accepted

As Sam Nead points out, it's not correct that every Seifert-fibred space is finitely covered by a product. The obstruction to this is the Euler number of the bundle. I strongly recommend Peter Scott's article 'The geometries of 3-manifolds' if you want to find out more about Seifert-fibred spaces. A scanned copy is available on his web page.

In the case when the base orbifold $O$ is torsion-free (and this is something that you always can assume by passing to a finite-sheeted cover---the point is that we only need to consider 'good' orbifolds, see Scott for details), the Euler number can be thought of as the obstruction to the short exact sequence

$1\to \mathbb{Z}\to\pi_1(M)\to\pi_1(O)\to 1$

splitting. Of course, this always happens when $\pi_1(O)$ is a free group, ie when $O$ is a surface with boundary.

This is why Hempel says you can assume that a finite-sheeted cover is a product---because he really cares about the case in which the JSJ decomposition is non-trivial. In this case, the Seifert-fibred pieces all have boundary, and so do indeed have a finite-sheeted cover that's a product.

Finally, it is true that you can prove residual finiteness of Seifert-fibred manifolds just from the short exact sequence. But care is needed! Deligne gave a famous example of a central extension of a lattice in a Lie group which is not itself residually finite.

For this reason, the proof is quite fiddly. The idea is to reduce to the case of a finite central extension of a surface group. You can in fact prove that any finite central extension of any residually finite one-relator group is residually finite, using Magnus's theorem that free groups are residually torsion-free nilpotent.

For full details, you could look for instance at this paper of Martino, in which he proves the stronger fact that Seifert-fibred 3-manifold groups are conjugacy separable.

share|cite|improve this answer
Thank you for the link to the Martino paper, that is a very nice proof indeed! – user641 Apr 19 '11 at 5:45

The proof of residual finiteness of the central cyclic extensions of surface groups is not hard: It suffices to consider the extension group with the presentation

$G=<a_1,b_1,...,a_g, b_g, t| [a_1,b_1]...[a_g,b_g]=t, [a_i,t]=1, [b_i,t]=1>$.

Now, kill all $a_i, b_i$ except for $a_1, b_1$. The quotient is the integer Heisenberg group $H=<a,b, t | [a,b]=t>$. The latter group is isomorphic to the group of integer 3x3 upper triangular matrices with 1's on the diagonal. We thus get a homomorphism $h: G\to H$ that is injective on the subgroup $<t>$. We also have the epimorphism $f: G\to F$, where $F$ is the surface group which kills the generator $t$. Now, take the product homomorphism $f\times h: G\to F\times H$ which is clearly injective. Since both $F$ and $H$ are residually finite, it follows that $G$ is residually finite as well. This argument also shows that $G$ is linear. It is an open problem (due to W.Thurston in 1980's) if the fundamental group of every compact 3-manifold $M$ is linear. (The hard case is when the JSJ-decomposition of $M$ is nontrivial, otherwise it follows from Perelman.)

share|cite|improve this answer

I think something is wrong here. If $M$ is the unit tangent bundle for a closed connected orientable surface $S$ of genus two (or larger) then $M$ is a Seifert fibered space, admitting $\widetilde{SL_2(R)}$ geometry. Following Scott (see the second to last paragraph of page 466 of his article "The geometries of 3-manifolds") the manifold $M$ cannot be finitely covered by a product $S^1 \times F$. This is because ${\rm Isom}\left(\widetilde{SL_2(R)}\right)$ does not contain closed hyperbolic surface groups.

You could cut the manifold $M$, given above, along a vertical torus. Then the base surface has boundary and so the unit tangent bundle has both $\widetilde{SL_2(R)}$ and $R \times H$ geometries. Because of the latter, there are covers that are products.

I don't have access to Hempel's article. If you know of a copy on-line please post a link here. When I searched for it I found only your other question:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.